Let $X$ and $Y$ be independent random variables such that $X \sim \text{Poisson}(\lambda \cdot c)$ and $Y \sim \text{Poisson}(\lambda \cdot (1-c))$, where $c$ is a real number in $[0, 1]$.

Is there an easy way of proving that $E\left[\frac{X}{X + Y} | X + Y > 0\right] = c$? Numerical computations I made indicate that this is true.

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    $\begingroup$ Google "Poisson thinning." $\endgroup$ – whuber Nov 14 '18 at 15:13
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    $\begingroup$ Aha! We can define Z := X + Y and see X as a thinning of Z with probability c. Then given any outcome of Z, conditional on Z, X is binomially distributed with parameters (Z, c), which has expected value c* Z. So for each Z!=0, E[X/Z | Z] = c. So E[X/Z | Z > 0] = c. Correct? $\endgroup$ – Christian Nov 14 '18 at 16:47

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